2015暑假多校联合---Friends（dfs枚举）

原题链接

[align=left]Problem Description[/align]
There are n people and m pairs of friends. For every pair of friends, they can choose to become online friends (communicating using online applications) or offline friends (mostly using face-to-face communication). However, everyone in these n people wants to have the same number of online and offline friends (i.e. If one person has x onine friends, he or she must have x offline friends too, but different people can have different number of online or offline friends). Please determine how many ways there are to satisfy their requirements.

[align=left]Input[/align]
The first line of the input is a single integer T (T=100), indicating the number of testcases.

For each testcase, the first line contains two integers n (1≤n≤8) and m (0≤m≤n(n−1)2), indicating the number of people and the number of pairs of friends, respectively. Each of the next m lines contains two numbers x and y, which mean x and y are friends. It is guaranteed that x≠y and every friend relationship will appear at most once.

[align=left]Output[/align]
For each testcase, print one number indicating the answer.

[align=left]Sample Input[/align]

2

3 3

1 2

2 3

3 1

4 4

1 2

2 3

3 4

4 1

[align=left]Sample Output[/align]

0

2

[align=left]Author[/align]
XJZX

[align=left]Source[/align]
2015 Multi-University Training Contest 2

[align=left]Recommend[/align]
wange2014

题意：输入n，m，n表示有n个人，m表示m对朋友关系，现在要使每个人的朋友划分为在线朋友和离线朋友，且在线朋友和离线朋友数量相等（一对朋友之间只能是在线朋友或者离线朋友），求方案数；

思路：用dfs深搜枚举每一条边（即每一对朋友关系），若能深搜进行完最后一条边，即当前边cnt==m+1 则ans++；

代码如下：

#include <iostream>
#include <algorithm>
#include <queue>
#include <vector>
#include <cstdio>
#include <cstring>
using namespace std;
int n,m,cnt,ans;
int c1[10],c2[10],d[10];
struct Node
{
int u,v;
}node[200];

void dfs(int i)
{
if(i-1==m)
{
ans++;
return ;
}
if(c1[node[i].u]&&c1[node[i].v])
{
c1[node[i].u]--;
c1[node[i].v]--;
dfs(i+1);
c1[node[i].u]++;
c1[node[i].v]++;
}
if(c2[node[i].u]&&c2[node[i].v])
{
c2[node[i].u]--;
c2[node[i].v]--;
dfs(i+1);
c2[node[i].u]++;
c2[node[i].v]++;
}
}

int main()
{
int T;
cin>>T;
while(T--)
{
cnt=0;
ans=0;
scanf("%d%d",&n,&m);
memset(node,0,sizeof(node));
memset(c1,0,sizeof(c1));
memset(c2,0,sizeof(c2));
memset(d,0,sizeof(d));
for(int i=1;i<=m;i++)
{
int u,v;
scanf("%d%d",&u,&v);
node[++cnt].u=u;
node[cnt].v=v;
d[u]++;
d[v]++;
}
int flag=0;
for(int i=1;i<=n;i++)
{
c1[i]=c2[i]=d[i]/2;
if(d[i]&1)
{
flag=1;
break;
}
}
if(flag)
{
puts("0");
continue;
}
dfs(1);
printf("%d\n",ans);
}
return 0;
}